Proof: use step 1, then use General
Retransposition to collect all the x terms in one bracket, and all the [x]
terms in another bracket.

Step 3: Such a bracket expression
equals, by Crosstransposition:

[[ [A] x ][ [B] [x] ]]C

Step 4: It therefore equals, by
Transposition:

[[ [A] C x ][ [B] C [x] ]]

Step 5: Let F(x) denote that
expression, then substitution yields

F(0) = [A]C

F(1) = [B]C

Therefore

F(x)=[[ F(0) x ][ F(1) [x] ]]

Call this the standard normal form.

1E. Complete deduction

Complete Deduction Theorem:

If the bracket form equation

F=G

is necessarily true, then the equation
is provable from the bracket axioms.

Proof is by induction on the number of
variables.

Case
0. IfF and G have no variables,
then the equation F=G is a bracket-arithmetic equation; these can be calculated
from the tables, which are a consequence of the bracket axioms. Therefore F=G
is provable from the axioms.

Case
N implies Case N+1. Suppose that all N-variable identities are provable
from the bracket axioms. Now suppose that F=G contains variable N+1; call it x.
Then these equations are provable:

Note that any proof thus constructed
will be equivalent to table look-up, as it will check all 2^n cases, for n =
number of variables; and thus take 2^n steps.

In complexity theory, finding exceptions is an NP-complete
problem.

1F. Incomplete re-entrance

Bracket forms are built of brackets
and juxtaposition of other bracket forms; a chain ultimately ending in the
void. Or does it? Could there be a form founded upon itself? Or even systems of
forms founded upon each other?

Consider the forms

A=[B]

B=[A]

This system, a ‘toggle’, has two
solutions:

A=0, B=1

A=1, B=0

Both make sense, but which is it?

Now consider a worse case; the form

L=[L]

L can equal neither 0 or 1.

Therefore Brownian bracket forms are
incomplete for re-entrance. To solve the equation, you need a third form.